Question

Totarlsulfexperimevt: Inorganic cations can be quantified by passing a salt solution throagh a cation-exchange columan in the ${\mathbf{OH}}^{\mathbf{-}}$form and titruting the relcasod H⁺with strong base. The moles of ${\mathbf{OH}}^{\mathbf{-}}$titrant equal the equivalenss of catioa charge in solution.

(a) What volume of 0.0231 M NaOH is needed to titrate the eluate when 10.00ml of $\mathbf{0}\mathbf{.}\mathbf{0458}{\mathbf{MKNO}}_{\mathbf{3}}$,have been loaded on a cation exchange columan in the H⁺foem?

(b) A 0.2692g sample of unknown salt is dissolved in demonized waler and loasedoato a cation exchange column in the H⁺form. The columin is rinsed with deionined water and the combincd loading and rinse solutions are titrated with 0.1396M KOH. A volume of 30.64mL is required to reach the endpoint. How many equivalents of cation are in the sample? Find the millieguivalents of cation per gram of sample (meq/g).

(c) The mass of substance containing one cquivalent is called the equivalent mas. If the cation has a +1 charge, the equivalent mass equals the molar mass. If the cation has a +2 charge, the equivalent mass is half the molar mass. What is the equivalent mass of the sample in (b)?

Short answer

1. The volume of NaOH required is = 19.83mL
2. Milli equivalents of cation that present in per gram of sample is 4.277 meq

The gine mass of sample is divided by calculated milli equivalent to give a equivalent mass of 0.2692 g sample.

Step-by-step solution

Find volume of 0.0231 M NaOH is needed to titrate the eluate

a)The volume of 0.0231M NaOH is needed to titrate the eluate form chromatography should be determined.

Ion-Exchange Chromatography:

Ion-Exchange Chromatography is separation technique, which is work in the principle of exchanging of ions based on attraction to the ion exchanger.

It contains two phases, one is stationary phase and another one is mobile phase.

In generally resins are act as a stationary phase, the positively charged ion exchangers attract solute anions and negatively charged ion exchangers are attract solute cations.

The higher polar eluent is passed through a column the exchangers are releases the solute and they will come out from the column.

In this process, the stationary phase (ion exchangers) is exchange the solute ions into eluent ions therefore it is called as lon-exchange chromatography.

The volume and concentration of unknown solution is determined by it is titrate with known volume and concentration solution.

The volume and concentration of unknown solution is required equivalent volume and concentration of known solution in the volumetric titration.

V₁M₁ = V₂M₂

Where,

V₁ is volume of known solution

N₁ is concentration of known solution

V₁ is volume of unknown solution

V₁is concentration of unknown solution

Mole:

The product of molarity of solution and volume of solution to give a mole of solute that present in the solution.

The required volume of NaOH to titrate the 10.00 mL of 0.0458 KNO3 is 19.83mL

Explanation of Solution

To determine the volume of 0.0231 M NaOH is needed to titrate the eluate form chromatography.

In ion-exchange chromatography, the mole of ions replaced is equal to the mole of ions present in the solute (sample). In cation-exchange column H⁺ is a Separator ion and the equivalent mole of H⁺ ion is calculated by titration to find the mole of cation present in the sample.

In the cation exchange chromatography${\mathrm{H}}^{-}$ is a common suppressor ion.

From the above statements, the mole of K⁺ ions are applied to the column is equal to the mole of H⁺ realised.

Therefore, the mole of NaOH required is equal to the mole of

$$\begin{gathered} \mathrm{NaOH}=0.0458\mathrm{M}\times 0.0100\mathrm{L} \\ =4.58\mathrm{M}{10}^{-4}\mathrm{mol} \end{gathered}$$

Hence, the volume of NaOH required is,

$$\begin{gathered} \mathrm{Molarity}=\frac{\mathrm{mole}}{\mathrm{Volume}} \\ \mathrm{Volume}=\frac{\mathrm{mole}}{\mathrm{Molarity}} \\ =0.0458\mathrm{M}\times 0.0100\mathrm{L} \\ =19.83\mathrm{mL} \end{gathered}$$

:Find themillieguivalents of cation per gram of sample

b) To determine the the milli equivalents of cation that present in per gram of sample.

In ion-exchange chromatography, the mole of ions replaced is equal to the mole of ions present in the solute (sample). In cation-exchange column H⁺is a Separator ion and the equivalent mole ofion is calculated by titration to find the mole of cation present in the sample.

In the cation exchange chromatography${\mathrm{OH}}^{-}$ is a common suppressor ion.

From the above statements, the mole equivalent of cation equal to the mole of${\mathrm{OH}}^{-}$ion

Therefore, the mole of NaOH required is equal to the mole ofrealised from the column.

Mole of NaOH=0.1396 M * 0.03064L

=4.27 mmol

The volume and concentration of NaOH are plugged in above equation to give the mole of NaOH required to titrate.

Hence, the milli equivalents of cation that present in per gram of sample is 4.277 meq

:Find equivalentmass of the sample

c)To determine the equivalent mass of 0.2692g sample.

In ion-exchange chromatography, the mole of ions replaced is equal to the mole of ions present in the solute (sample). In cation-exchange column H⁺ is a Separator ion and the equivalent mole ofH⁺ ion is calculated by titration to find the mole of cation present in the sample.

In the cation exchange chromatography${\mathrm{H}}^{-}$is a common suppressor ion.

From the above statements, the mole equivalent of cation equal to the mole of${\mathrm{OH}}^{-}$ion

Therefore, the mole of NaOH required is equal to the mole of realised from the column.

Mole of NaOH=0.1396 M * 0.03064 L

=4.27 mmol

The volume and concentration of NaOH are plugged in above equation to give the mole of NaOH required to titrate.

Hence, the milli equivalents of cation that present in per gram of sample is 4.277meq

The equivalent mass id given by the mass of substance that contains one equivalent.

The charge of give sample is +1 so the equivalent is,

$$\begin{gathered} =\frac{0.269\mathrm{g}}{4.277\times {10}^{-3}\mathrm{eq}} \\ =62.94\mathrm{g}/\mathrm{eq} \end{gathered}$$

The gine mass of sample is divided by calculated milli equivalent to give a equivalent mass of 0.2692 g sample.